$\sum\limits_{n=1}^{\infty } \dfrac{(x-4)^n}{n\cdot(-2)^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $2 \le x < 6$ (Choice B) B $0<x \le 8$ (Choice C) C $0 \le x < 8$ (Choice D) D $2<x \le6$
Solution: We use the ratio test. For $x\neq 4$, let $a_n= \dfrac{(x-4)^n}{n\cdot(-2)^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x-4}{2}\right| $ The series converges when $\left|\dfrac{x-4}{2}\right| <1$, which is when $2<x<6$. Now let's check the endpoints, $x=2$ and $x=6$. Letting $x=2$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(2-4)^n}{n\cdot(-2)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(-2)^n}{n\cdot(-2)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{1}{n} \end{aligned}$ This is the harmonic series, which is known to diverge. Letting $x=6$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(6-4)^n}{n\cdot(-2)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{2^n}{n\cdot(-2)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot2^n}{n\cdot 2^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n} \end{aligned}$ This is the alternating harmonic series, which is known to converge. In conclusion, the interval of convergence is $2<x \le6$.